As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are $30^{o}$ and $45^{o}$ If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use $\sqrt{3}=1.732$]

Given: Height of the light house$=100\ m$,  the angles of depression of two ships are $30^{o}$ and $45^{o}$ and one ship is exactly behind the other on the same side of the light house.

To do: To find the distance between the two ships.

Solution:

Let ships are at distance $x$ from each other 

In $\vartriangle APO$

$tan45^{o}=\frac{100}{y}=1$

$\Rightarrow y=100\ m$                          ...$( i)$

In $\vartriangle POB$,

$tan30^{o}=\frac{OP}{OB}=\frac{100}{x+y}=\frac{1}{\sqrt{3}}$

$\Rightarrow x+y=100\sqrt{3} $                

On subtituting $y=100\ m$ from $( i)$,

$x+100=100\sqrt{3}$

$\Rightarrow x=100(\sqrt{3}-1)$

$\Rightarrow x=100(1.732-1)$

$\Rightarrow x=73.2\ m$

$\therefore$ Ships are $73.2\ meter$ apart.
 

Updated on: 10-Oct-2022

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